Fourier Series and Partial Differential Equations Lecture Notes

Chapter 2. Fourier series 13
we must have
a0 = 1
π
f(x)dx. (2.22)
π −π
Thus if equation (2.19) is true, then we know a0.
Note that due to the properties of periodic functions, we could use 2π
0 f(x)dx instead
of π
−πf(x)dx in the preceding. More generally, we could integrate over any interval of
length 2π.
Lemma 2.1 Let n,m ∈ N\0. We then have the following equalities:
π
sin(mx)cos(nx)dx = 0, (2.23)
−π
π
sin(mx)sin(nx)dx = πδnm, (2.24)
−π
π
cos(mx)cos(nx)dx = πδnm, (2.25)
−π
where δnm is the Kronecker delta defined by
δnm =
0 n = m,
1 n = m.
(2.26)
Proof. Equation (2.23) is trivial as sin(mx)cos(nx) is odd. For equation (2.24) we compute,
for n = m,
π
sin(mx)sin(nx)dx =
−π
1
π
[−cos{(m+n)x}+cos{(m−n)x}] dx,
2 −π
= 1
−sin{(m+n)x}
+
2 m+n
sin{(m−n)x}
π ,
m−n −π
= 0. (2.27)
If n = m we have
π π
1−cos(2nx)
sin(nx)sin(mx)dx =
2
−π
−π
Similar computations yield equation (2.25).
dx = 1
x−
2
sin(2nx)
π = π. (2.28)
2n −π
Thus, to find an and bn, we assume equation (2.19) is true. Multiplying both sides by
cos(mx) and integrating term-wise gives
π
f(x)cos(mx)dx =
−π
1
2 a0
π
cos(mx)dx
−π
∞
π
+ cos(nx)cos(mx)dx
+
n=1
∞
n=1
an
bn
−π
π
sin(nx)cos(mx)dx. (2.29)
−π